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A car is driven east for a distance of 54 km, then north for 32 km, and then in a direction 28° east of north for 27 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

A car is driven east for a distance of 54 km, then north for 32 km, and then in a direction 28° east of north for 27 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Assumption:

Let us assume that the vector\overrightarrow{a} represents the eastward motion of the car with a magnitude of , the northward motion of the car is given by vector\overrightarrow{b} with a magnitude of 32 km while the east of north motion by vector\overrightarrow{c} with a magnitude of
27 km.
The figure below shows the vector diagram of the Car’s motion:
236-1524_1.PNG
In vector\overrightarrow{a} , there is no vertical component but only the horizontal component along the east with magnitude 54 km. Therefore vector\overrightarrow{a} is:
\overrightarrow{a} = 54\ km\ \widehat{i}
For vector \overrightarrow{b}, there is no horizontal component as it describes the northward motion of the car, along the unit vector \widehat{j}.
The magnitude of the vertical component of the car is 32 km, therefore vector\overrightarrow{b} can be written as:
\overrightarrow{b} = 32\ km\ \widehat{j}
From the vector diagram above, it is clear that the horizontal component of vector\overrightarrow{c} points in the direction of unit vector\widehat{i} . Therefore the horizontal component of vector\overrightarrow{c} is .
27 km sin (28°)
The vertical component of the vector\overrightarrow{c} points northward, along the unit vector \widehat{j}, therefore the component is.27 km cos(28°) .
Therefore vector \overrightarrow{c}is:
236-1637_2.PNG
The magnitude of total vertical displacement (say sv) of the car along the north is the sum of vertical component of vector\overrightarrow{a} , the vertical component of vector\overrightarrow{b} and the vertical component of vector \overrightarrow{c}.
Since vector \overrightarrow{a}does not have a vertical component, we add the vertical component of vector\overrightarrow{b} and \overrightarrow{c}, to obtain the magnitude of displacement sv as:
236-132_3.PNG
The magnitude of total horizontal displacement (say sh) of the car along the east is the sum of horizontal component of vector \overrightarrow{a}, the horizontal component of vector\overrightarrow{b} and the horizontal component of vector \overrightarrow{c}.
Since vector \overrightarrow{b}does not have a horizontal component, we add the horizontal component of vector\overrightarrow{a} and \overrightarrow{c}, to obtain the magnitude of displacement sh as:
236-1791_4.PNG
The magnitude of net displacement (say s) of the car can be calculated using the magnitude in equation (1) and (2) as:
236-1871_5.PNG
Rounding off to two significant figures
s = 87 km
Therefore the magnitude of displacement of the car during his journey is 87 km .

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